解:$(1)M=(2x^2+3xy + 2y)-2(x^2-xy + x-\frac {1}{2})$
$ =2x^2+3xy + 2y-2x^2+2xy - 2x + 1$
$ =5xy + 2y-2x + 1$
$ $当$x=\frac {1}{5},$$y = - 1$时,
$ $原式$=5×\frac {1}{5}×(-1)+2×(-1)-2×\frac {1}{5}+1$
$ =-1-2-\frac {2}{5}+1$
$ =-2-\frac {2}{5}$
$ =-2\frac {2}{5}$
$ (2)$因为$M = 5xy + 2y-2x + 1=(5y - 2)x + 2y + 1,$多项式$M$的值与字母$x$的取值无关,
$ $所以$5y - 2 = 0,$
$ $解得$y=\frac {2}{5}$